Longest palindromic String formed using concatenation of given strings in any order

Given an array of strings arr[] of the same length, the task is to find the longest palindromic string that can be made using the concatenation of strings in any order.

Examples:

Input: arr[] = {“aba”, “aba”}
Output: abaaba

Input: arr[] = {“abc”, “dba”, “kop”, “cba”, “abd”}
Output: abcdbaabdcba

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

Find all the pairs of strings which are reverse of each other and store them in two different arrays pair1[] and pair2 separately and delete those pairs from the original array.
Find any palindromic string s1 in the array.
Join all the strings together of the array pair1[] into s2
Join all the strings together of the array pair2[] in reverse order into s3
Concatenate the strings s2 + s1 + s3 together to get longest palindromic string.

Below is the implementation of the above approach.

C++

// C++ implementation to find the longest 
// palindromic String formed using 
// concatenation of given strings in any order 
  
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to find the longest palindromic 
// from given array of strings 
void longestPalindrome(string a[], 
                       int n) 
{ 
    string pair1[n]; 
    string pair2[n]; 
    int r = 0; 
  
    // Loop to find the pair of strings 
    // which are reverse of each other 
    for (int i = 0; i < n; i++) { 
        string s = a[i]; 
        reverse(s.begin(), s.end()); 
        for (int j = i + 1; j < n; j++) { 
            if (a[i] != "" && a[j] != "") { 
                if (s == a[j]) { 
                    pair1[r] = a[i]; 
                    pair2[r++] = a[j]; 
                    a[i] = ""; 
                    a[j] = ""; 
                    break; 
                } 
            } 
        } 
    } 
    string s1 = ""; 
  
    // Loop to find if any palindromic 
    // string is still left in the array 
    for (int i = 0; i < n; i++) { 
        string s = a[i]; 
        reverse(a[i].begin(), a[i].end()); 
        if (a[i] != "") { 
            if (a[i] == s) { 
                s1 = a[i]; 
                break; 
            } 
        } 
    } 
    string ans = ""; 
  
    // Update the answer with 
    // all strings of pair1 
    for (int i = 0; i < r; i++) { 
        ans = ans + pair1[i]; 
    } 
    // Update the answer with 
    // palindromic string s1 
    if (s1 != "") { 
        ans = ans + s1; 
    } 
    // Update the answer with 
    // all strings of pair2 
    for (int j = r - 1; j >= 0; j--) { 
        ans = ans + pair2[j]; 
    } 
    cout << ans << endl; 
} 
  
// Driver Code 
int main() 
{ 
    string a1[2] = { "aba", "aba" }; 
    int n1 = sizeof(a1) / sizeof(a1[0]); 
    longestPalindrome(a1, n1); 
  
    string a2[5] = { "abc", "dba", "kop", 
                     "abd", "cba" }; 
    int n2 = sizeof(a2) / sizeof(a2[0]); 
    longestPalindrome(a2, n2); 
} 

Java

// Java implementation to find the longest 
// palindromic String formed using 
// concatenation of given Strings in any order 
class GFG 
{ 
  
// Function to find the longest palindromic 
// from given array of Strings 
static void longestPalindrome(String a[], 
                            int n) 
{ 
    String []pair1 = new String[n]; 
    String []pair2 = new String[n]; 
    int r = 0; 
  
    // Loop to find the pair of Strings 
    // which are reverse of each other 
    for (int i = 0; i < n; i++) 
    { 
        String s = a[i]; 
        s = reverse(s); 
        for (int j = i + 1; j < n; j++)  
        { 
            if (a[i] != "" && a[j] != "") 
            { 
                if (s.equals(a[j]))  
                { 
                    pair1[r] = a[i]; 
                    pair2[r++] = a[j]; 
                    a[i] = ""; 
                    a[j] = ""; 
                    break; 
                } 
            } 
        } 
    } 
    String s1 = ""; 
  
    // Loop to find if any palindromic 
    // String is still left in the array 
    for (int i = 0; i < n; i++)  
    { 
        String s = a[i]; 
        a[i] = reverse(a[i]); 
        if (a[i] != "")  
        { 
            if (a[i].equals(s)) 
            { 
                s1 = a[i]; 
                break; 
            } 
        } 
    } 
    String ans = ""; 
  
    // Update the answer with 
    // all Strings of pair1 
    for (int i = 0; i < r; i++) 
    { 
        ans = ans + pair1[i]; 
    } 
      
    // Update the answer with 
    // palindromic String s1 
    if (s1 != "") 
    { 
        ans = ans + s1; 
    } 
    // Update the answer with 
    // all Strings of pair2 
    for (int j = r - 1; j >= 0; j--)  
    { 
        ans = ans + pair2[j]; 
    } 
    System.out.print(ans +"\n"); 
} 
static String reverse(String input)  
{ 
    char[] a = input.toCharArray(); 
    int l, r = a.length - 1; 
    for (l = 0; l < r; l++, r--) 
    { 
        char temp = a[l]; 
        a[l] = a[r]; 
        a[r] = temp; 
    } 
    return String.valueOf(a); 
} 
  
// Driver Code 
public static void main(String[] args) 
{ 
    String []a1 = { "aba", "aba" }; 
    int n1 = a1.length; 
    longestPalindrome(a1, n1); 
  
    String []a2 = { "abc", "dba", "kop", 
                    "abd", "cba" }; 
    int n2 = a2.length; 
    longestPalindrome(a2, n2); 
} 
} 
  
// This code is contributed by Rajput-Ji 

Python3

# Python3 implementation to find the longest 
# palindromic String formed using 
# concatenation of given strings in any order 
  
# Function to find the longest palindromic 
# from given array of strings 
def longestPalindrome(a, n): 
    pair1 = [0]*n 
    pair2 = [0]*n 
    r = 0
  
    # Loop to find the pair of strings 
    # which are reverse of each other 
    for i in range(n): 
        s = a[i] 
        s = s[::-1] 
        for j in range(i + 1, n): 
            if (a[i] != "" and a[j] != ""): 
                if (s == a[j]): 
                    pair1[r] = a[i] 
                    pair2[r] = a[j] 
                    r += 1
                    a[i] = "" 
                    a[j] = "" 
                    break
  
    s1 = "" 
  
    # Loop to find if any palindromic 
    # is still left in the array 
    for i in range(n): 
        s = a[i] 
        a[i] = a[i][::-1] 
        if (a[i] != ""): 
            if (a[i] == s): 
                s1 = a[i] 
                break
  
    ans = "" 
  
    # Update the answer with 
    # all strings of pair1 
    for i in range(r): 
        ans = ans + pair1[i] 
      
    # Update the answer with 
    # palindromic s1 
    if (s1 != ""): 
        ans = ans + s1 
      
    # Update the answer with 
    # all strings of pair2 
    for j in range(r - 1, -1, -1): 
        ans = ans + pair2[j] 
    print(ans) 
  
# Driver Code 
a1 = ["aba", "aba"] 
n1 = len(a1) 
longestPalindrome(a1, n1) 
  
a2 = ["abc", "dba", "kop","abd", "cba"] 
n2 = len(a2) 
longestPalindrome(a2, n2) 
  
# This code is contributed by mohit kumar 29 

C#

// C# implementation to find the longest 
// palindromic String formed using 
// concatenation of given Strings in any order 
using System; 
  
class GFG 
{ 
   
// Function to find the longest palindromic 
// from given array of Strings 
static void longestPalindrome(String []a, 
                            int n) 
{ 
    String []pair1 = new String[n]; 
    String []pair2 = new String[n]; 
    int r = 0; 
   
    // Loop to find the pair of Strings 
    // which are reverse of each other 
    for (int i = 0; i < n; i++) 
    { 
        String s = a[i]; 
        s = reverse(s); 
        for (int j = i + 1; j < n; j++)  
        { 
            if (a[i] != "" && a[j] != "") 
            { 
                if (s.Equals(a[j]))  
                { 
                    pair1[r] = a[i]; 
                    pair2[r++] = a[j]; 
                    a[i] = ""; 
                    a[j] = ""; 
                    break; 
                } 
            } 
        } 
    } 
    String s1 = ""; 
   
    // Loop to find if any palindromic 
    // String is still left in the array 
    for (int i = 0; i < n; i++)  
    { 
        String s = a[i]; 
        a[i] = reverse(a[i]); 
        if (a[i] != "")  
        { 
            if (a[i].Equals(s)) 
            { 
                s1 = a[i]; 
                break; 
            } 
        } 
    } 
    String ans = ""; 
   
    // Update the answer with 
    // all Strings of pair1 
    for (int i = 0; i < r; i++) 
    { 
        ans = ans + pair1[i]; 
    } 
       
    // Update the answer with 
    // palindromic String s1 
    if (s1 != "") 
    { 
        ans = ans + s1; 
    } 
    // Update the answer with 
    // all Strings of pair2 
    for (int j = r - 1; j >= 0; j--)  
    { 
        ans = ans + pair2[j]; 
    } 
    Console.Write(ans +"\n"); 
} 
static String reverse(String input)  
{ 
    char[] a = input.ToCharArray(); 
    int l, r = a.Length - 1; 
    for (l = 0; l < r; l++, r--) 
    { 
        char temp = a[l]; 
        a[l] = a[r]; 
        a[r] = temp; 
    } 
    return String.Join("",a); 
} 
   
// Driver Code 
public static void Main(String[] args) 
{ 
    String []a1 = { "aba", "aba" }; 
    int n1 = a1.Length; 
    longestPalindrome(a1, n1); 
   
    String []a2 = { "abc", "dba", "kop", 
                    "abd", "cba" }; 
    int n2 = a2.Length; 
    longestPalindrome(a2, n2); 
} 
} 
  
// This code is contributed by 29AjayKumar 

Output:

abaaba
abcdbaabdcba

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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